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Arithmetic Progressions — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsArithmetic Progressions3 Marks
Question
Two arithmetic progression have the same common difference. The difference between their $100^{th}$ terms is $100,$ What is the difference between their $1000^{th}$ terms$?$
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Answer

Let the two $A.P.$ is be $a_1, a_2, a_3, \ldots$. and $b_1, b_2, b 3, \ldots$.
$a_n=a_1+(n-1) d \text { and } b_n=b_1+(n-1) d$
Since common difference of two equations is same given difference between $100^{\text {th }}$ terms is $100$
$ a_{100}-b_{100}=100 $
$ a_1+(99) d-b_1-99 d=100 $
$ a_1-b_1=100 \ldots . \text { (i) }$
Difference between. $1000$ th terms is
$ a_{1000}-b_{1000}=a_1+(1000-1) d-\left(b_1+(1000-1) d\right) $
$ =a_1+999 d-b_1-999 d $
$ =a_1-b_1 $
$ =100(\text { from }(1))$
$\therefore$ Hence difference between $1000^{\text {th }}$ terms of two $A.P.$ is $100 .$

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